Question: Simplify; express your answer in exponential form. Assume $n\neq 0, k\neq 0$. $\dfrac{{(n^{3})^{-1}}}{{(n^{-1}k^{-2})^{-5}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{3}}$ to the exponent ${-1}$ . Now ${3 \times -1 = -3}$ , so ${(n^{3})^{-1} = n^{-3}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-1}k^{-2})^{-5} = (n^{-1})^{-5}(k^{-2})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{3})^{-1}}}{{(n^{-1}k^{-2})^{-5}}} = \dfrac{{n^{-3}}}{{n^{5}k^{10}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-3}}}{{n^{5}k^{10}}} = \dfrac{{n^{-3}}}{{n^{5}}} \cdot \dfrac{{1}}{{k^{10}}} = n^{{-3} - {5}} \cdot k^{- {10}} = n^{-8}k^{-10}$.